Question

我有一个数据框，其中一列（'ProcessingDATE'）是日期时间格式。我想创建另一列（“报告日期”），如果处理日期是星期一，则从中减去3天，最后以星期五结束；否则减去1天。

我使用python的时间很短，因此对如何编写它没有太多的了解。我的想法是用"provider": { "@type": "LocalBusiness", "@id": "https://example.com/#LocalBusiness" }，cell = Monday编写一个for循环； then = datetime.datetime.today() – datetime.timedelta(days=3)

else = datetime.datetime.today() – datetime.timedelta(days=1)

Answer 1

希望这会有所帮助，

import json
import requests
from requests.exceptions import HTTPError
host  = "http://httpbin.org/post"
host = "http://webservicehere.azurewebsites.net"
f = open('hard_code.xml', 'rb').read()
try:
  response = requests.head(host,data=f)
  response.encoding = "utf-8"
  print("Encoding " + response.encoding)
  #response = requests.get('https://api.github.com/search/repositories',
  #params={'q': 'requests+language:python'})
  response.raise_for_status()
  print(response.status_code)
  print(response.text)
  if 'json' in response.headers.get('Content-Type'):
    json_response = response.json()
    print(json_response)
  else: 
    print("Response not in JSon form")
except HTTPError as http_err:
    print(f'HTTP error occurred: {http_err}')  # Python 3.6

except Exception as err:
    print(f'Other error occurred: {err}')  # Python 3.6

else:
    print('Success!')

以上内容也可以写成

from datetime import timedelta

if DDA_compamy['ProcessingDATE'].weekday() == 4:  #Condition to check if it is friday
    DDA_compamy['Report Date']=DDA_compamy['ProcessingDATE'] - timedelta(days=3) # if friday subtracting 3 days
else:
    DDA_compamy['Report Date']=DDA_compamy['ProcessingDATE'] - timedelta(days=1) #Else one day from the date is subtracted

Answer 2

使用pandas.Series.dt.weekday和一些逻辑：

import pandas as pd
df = pd.DataFrame({'ProcessingDATE':pd.date_range('2019-04-01', '2019-04-27')})
df1 = df.copy()
mask = df1['ProcessingDATE'].dt.weekday == 0
df.loc[mask, 'ProcessingDATE'] = df1['ProcessingDATE'] - pd.to_timedelta('3 days')
df.loc[~mask, 'ProcessingDATE'] = df1['ProcessingDATE'] - pd.to_timedelta('1 days')

输出：

   ProcessingDATE
0      2019-03-29
1      2019-04-01
2      2019-04-02
3      2019-04-03
4      2019-04-04
5      2019-04-05
6      2019-04-06
7      2019-04-05
8      2019-04-08
9      2019-04-09
10     2019-04-10
11     2019-04-11
12     2019-04-12
13     2019-04-13
14     2019-04-12
15     2019-04-15
16     2019-04-16
17     2019-04-17
18     2019-04-18
19     2019-04-19
20     2019-04-20
21     2019-04-19
22     2019-04-22
23     2019-04-23
24     2019-04-24
25     2019-04-25
26     2019-04-26

从日期时间列中减去日期

2 个答案: